What is the formulae for higher-order finite differences in several variables? Wikipedia's article on finite difference gives the first-order approximations in several variables:
$f_{xy}(x,y)\approx \frac{f(x+h,y+k)-f(x+h,y-k)-f(x-h,y+k)+f(x-h,y-k)}{4hk}$
And this source (Eq. 13) provides the next-order approximations in one variable:
$f''(x) \approx \frac{-f(x+2h)+16f(x+h)-30f(x)+16f(x-h)-f(x-2h)}{12h^2}$
I'm looking for the equation for the second-order approximation in several variables. Third-order approximations or higher, if available, would be even better. What are they, or where can I find them?
The form of the first equation is very suggestive when compared to the first-order approximation in one variable, and seems to suggest it's easy to go from that formula to the one involving multiple variables. In particular all I have to do to go from this equation:
$f'(x) \approx \frac{f(x+h)-f(x-h)}{2h}$
to the first equation in this question is to assume that the two $f$ terms in the numerator have a corresponding y-term, subtract those, and then multiply the denominator by $2k$. However I'm not confident enough to attempt this for the second derivative, especially since there are five terms in the numerator (does this mean the formula I'm looking for will have $25$ terms in the numerator?).
 A: (Answering my own question since I figured it out)
This is actually discussed in section 3 of one of the sources quoted in the question. The idea is to take the approximation to the first derivative and apply it twice. So this indeed leads to a lot of terms.
However, there aren't actually 25 terms because there are only five terms in $f''(x)$. If one is after the formula for the second derivative, one needs to start with the formula for the first derivative. The second-order formula for the first derivative is:
$f'(x) \approx \frac{1}{12}f(x-2h) -\frac{2}{3}f(x-h) + \frac{2}{3}f(x+h) - \frac{1}{12}f(x+2h)$
i.e. there are only four terms. The second-order formula for the second derivative therefore has $16 = 4^2$ terms. In the notation of the source, the coefficients are
$C_{2,0} = -\frac{1}{12h}$, $C_{1,0} = \frac{2}{3h}$, $C_{-1,0} = -\frac{2}{3h}$, $C_{-2,0} = \frac{1}{12h}$
The coefficients for the $C_{0,2}$ terms etc can be gotten similarly, but with $k$ instead of $h$ since it's $y$ being modified now, and the starting equation is:
$f'(x) \approx \frac{1}{12}f(y-2k) -\frac{2}{3}f(y-k) + \frac{2}{3}f(y+k) - \frac{1}{12}f(y+2k)$
The coefficients of the second-order formula for the second derivative can be gotten by multiplying these, e.g.
$C_{-2,-2} = C_{-2,0} \times C_{0,-2} = \frac{1}{144}$
Translating back into explicit form, this means that the coefficient of the $f(x-2h,y-2k)$ term is $1/144$. Adding all sixteen terms yields the desired formula.
